Excel Sheet Column Number

easy math string

Problem

Given a string columnTitle (the column label in an Excel sheet), return its corresponding column number.

InputcolumnTitle = "ZY"

Output701

Bijective base 26: result starts at 0; for each char, result = result*26 + (c - 'A' + 1).

columnTitle

def title_to_number(s):
    n = 0
    for c in s:
        n = n * 26 + (ord(c) - ord('A') + 1)
    return n

function titleToNumber(s) {
  let n = 0;
  for (const c of s) {
    n = n * 26 + (c.charCodeAt(0) - 64);
  }
  return n;
}

class Solution {
    public int titleToNumber(String s) {
        int n = 0;
        for (int i = 0; i < s.length(); i++) {
            n = n * 26 + (s.charAt(i) - 'A' + 1);
        }
        return n;
    }
}

int titleToNumber(string s) {
    int n = 0;
    for (char c : s) {
        n = n * 26 + (c - 'A' + 1);
    }
    return n;
}

Explanation

Excel column labels are just numbers written in base 26, where the digits are the letters A-Z. We decode the string the same way you read a normal number: process letters left to right, building up the value as we go.

Each letter is worth its position in the alphabet: A=1, B=2, ..., Z=26. In code that is ord(c) - ord('A') + 1. Notice there is no zero — this is a bijective base-26 system, so A is 1 rather than 0.

For every letter we do n = n * 26 + value. Multiplying by 26 shifts the previous total one place to the left (just like multiplying by 10 in decimal), then we add the new letter's value.

Example: "ZY". Start n = 0. Z is 26, so n = 0 * 26 + 26 = 26. Then Y is 25, so n = 26 * 26 + 25 = 701. The answer is 701.

We touch each character once, so the work is proportional to the length of the label.

Time: O(|s|) Space: O(1)